Why doesn't induction work on $\mathbb Q$?

Question

I'm trying to better understand induction. Given that there's a bijection between $\mathbb N$ and $\mathbb Q^+$ (positive rationals), why can't I define the successor relationship on $\mathbb Q$ based on the $\mathbb Q \rightarrow \mathbb N$ mapping. For argument's sake, say that I want to prove that $\forall x \in \mathbb Q^+$, $x$ $\ge 0$. I show this for the first $\mathbb Q$ (in whatever order I enumerate $\mathbb Q$), then show that if it's true for the $k$th positive $\mathbb Q$, it's also true for the $(k+1)$th positive $\mathbb Q$.

Why am I not able to conclude using induction that it's true for all $\mathbb Q^+$? Given that I have a bijection, why doesn't the same well-ordering argument work? i.e., if some set of $x \in\mathbb Q^+< 0$, then there's the smallest one based on the bijection with $\mathbb N$. I'm missing something fundamental here, although I do understand that between any two rationals, there is an infinite in between.

Answer 1

Yes you can. All you would need to do for your steps is: Show if your condition to be satisfied is is $P$, then $P(r)$ holds for all $r \in \Bbb{Q}$. That is, show

1. $P(0)$ holds
2. If $P(x)$ holds, then $P(-x)$ holds. Here of course, $x\in\Bbb{Q}$.
3. If $P(\frac{p}{q})$ holds, then $P(\frac{p+1}{q})$ holds, Here of course, $p,q\in\Bbb{Z}$.
4. If $P(\frac{p}{q})$ holds, then $P(\frac{p}{q+1})$ holds, and here of course assume $q\neq -1$ and again, $p,q\in\Bbb{Z}$.

this should suffice.

Answer 2

I don't believe there to be anything prohibiting you from inducting on the rational numbers as you do on the natural numbers—at least, in theory. However, in practice, the reason I think we don't often see this is that the entire usefulness of induction over the natural numbers is the constant recurrence relation between a natural number and its successor, which the rational numbers lack. Because there is no straightforward, canonical well-ordering of the rational numbers like there is for the natural numbers, it becomes hard to carry out the inductive step, "assuming $P(q_{n})$, then $P(q_{n+1})$. All that to say: you could define a well-ordering of $\mathbb{Q}$ and perform induction just as you do on $\mathbb{N}$, but, how you actually go about proving anything using this method would be difficult because of lacking a good ordering.

Perhaps, this question (induction on the rational numbers) may be of interest to you, or if you want an even more exotic case, this question (considering induction on the real numbers). Hope this helps!

Edit: I now see the comments on your original question that more or less seem to better explain what I was trying to convey, so I apologize for the late response, but hopefully something in my answer is of some help.

Answer 3

Assuming the axiom of choice, every set can be well-ordered. Assume $\prec$ is a well order on $\mathbb{Q}$. This means that you actually have a successor on $\mathbb{Q}$, by using a property of minimum ("every non empty subset has a minimal element"). $$S(q) := \min\{s \in \mathbb{Q} : s \succ q\}$$

If you look for the proof of Induction Theorem, the only thing you need is this minimum property.

The other thing is why it is useless. On $\mathbb{N}$ induction is helpful when you have clear connection between k-th and k+1-st step. With such an abstract successor on $\mathbb{Q}$ it would not be helpful at all.

Answer 4

Bear with me.

Suppose you wanted to prove that all natural numbers have a unique prime factorization and you want to do it by induction.

So you set up you base cases. And then you attempt your induction step. So you need to show that if $k$ has a unique prime factorization $\prod p_i^{a_i}$ then $k+1= \prod p_i^{a_i}+1$ has a unique prime factorization.

Well, how the #@\$%! are you going to do that?! There is utterly nothing about $k=\prod p_i^{a_1}$ that can possibly lead to concluding $k+1 =\prod q_i^{b_i}$ has a unique prime factorization.

And that is the impracticality of induction on $\mathbb Q$.

If you had an bijection $f:\mathbb N \to \mathbb Q$, then yes, you can show that a proposition holds for base case $P(f(0))$. But then if you can show that $P(f(k))$ must imply that $P(f(k+1))$ holds then, yes, you are done. You have proven by induction that $P$ holds for all $\mathbb Q$.

So you can do induction on $\mathbb Q$.

But look at your induction step. If the property holds for some rational $q = f(k)$. You have to prove that there is a logical reason it should hold for $f(k+1)$. But $f$'s only significance is as function to map one natural number to a rational. There is no reason (and lots of reasons why not) that this property $P$ which is about unordered normal rational numbers should have any structural similarity with some abstract external ordering of the naturals.

In fact I can't think of any examples (except those that are specifically about the ordering itself).